The short answer
No, a general multivector is not graded. Only a portion of the elements of the algebra are assigned grades. Yes blades are among the elements assigned grades. A blade represents a subspace of $V$, and the grade of that blade is the dimension of the subspace the blade represents.
Everything I'm about to say is with regards to an $n$ dimensional quadratic space $V$ with a nondegenerate form.
Hopefully you are familiar with the construction where an orthonormal basis $\{e_i\mid 1\leq i\leq n\}$ for $V$ is used to construct a basis for the Clifford algebra.
The basis for the Clifford algebra is made up from these sets:
$S_0=\{1\}$
$S_1=\{e_i\mid 1\leq i\leq n\}$
$S_2=\{e_ie_j\mid 1\leq i< j\leq n \}$
$S_3=\{e_ie_je_k\mid 1\leq i< j<k\leq n \}$
...
$S_n=\{e_1e_2\cdots e_n\}$
The basis for the Clifford algebra we are talking about is $\cup_{j=1}^n S_j$.
What a grade is:
Everything in the vector space $\langle S_i \rangle$ spanned by $S_i$ is said to have grade $i$ (excepting 0 for the time being). Things which are sums of elements of different grades do not have any grade defined for them, and they are said to have mixed grade. Grade only applies to things in these special subspaces. The element 0 lies in all of these sets, so one needs to understand 0 as being an exceptional element. (I'm not exactly sure what most references do concerning 0.)
It isn't hard to show that any product of $k$ vectors (vectors, elements of $V$ lie in $\langle S_1 \rangle$) lies in $\langle S_k\rangle$. It turns out that the product is zero exactly when the set of vectors you are multiplying is linearly dependent. Blades, being products like this, do always have a grade.
What grade roughly means:
For a blade $b_1\cdots b_k$ of $k$ vectors from $V$ made up of linearly independent vectors $b_i$", you get an element of grade $k$. Since this element is usually interpreted as "the subspace of $V$ generated by the $b_i$, you can see that the grade is coinciding with the dimension of this subspace.
When the vectors are not LI, things are different. Had the $b_i$ been linearly dependent, the product $b_1\cdots b_k$ would be zero, but the dimension of the span of the $b_i$ could be anything less than $k$, so you can see why there are problems assigning 0 a fixed grade to make the system of "grade is the same as dimension of span" work universally.
The interesting thing is that multiplication plays well with grades. I think the path most often taken is to just define the grade of 0 to be 0, and then declare "the product of a grade $k$ vector with a grade $j$ vector has grade $j+k$ or grade $0$ according to whether the product is zero or not."
Best Answer
For the first decomposition, since $ x \cdot B $ is a vector, $ x \wedge ( x \cdot B ) $ must be a bivector (grade 2). The product of the bivector $ A $ with another bivector is a multivector with grades that may include 0, 2, 4, and the dot product of those two is the grade-0 selection. That is $$\begin{aligned} A \cdot \left( { x \wedge (x \cdot B) } \right) &= \left\langle{{ A \left( { x \wedge (x \cdot B) } \right) }}\right\rangle \\ &= \left\langle{{ A \left( { x (x \cdot B) - x \cdot (x \cdot B) } \right) }}\right\rangle.\end{aligned}$$ Here we used $ x y = x \cdot y + x \wedge y $, which leaves us a with a scalar factor $ x \cdot (x\cdot B) $, that has no contribution to the grade-0 selection, which leaves us with $$\begin{aligned} A \cdot \left( { x \wedge (x \cdot B) } \right) &= \left\langle{{ A x (x \cdot B) }}\right\rangle \\ &= \left\langle{{ (A \cdot x) (x \cdot B) }}\right\rangle \\ &= \left\langle{{ (A \cdot x) (x B - x \wedge B) }}\right\rangle \\ &= \left\langle{{ (A \cdot x) x B }}\right\rangle.\end{aligned}$$ Here we made use of the fact that $ A x = A \cdot x + A \wedge x $, a vector and trivector, of which only the vector component contributes to the scalar selection. Then we are free to rewrite $ x \cdot B $ in terms of $ x B $ and a trivector that also has no contritution to the scalar selection. The final result follows from the fact that reversion of a scalar leaves it unchanged, so $$\begin{aligned} \left\langle{{ (A \cdot x) x B }}\right\rangle &= \left\langle{{ B x (x \cdot A) }}\right\rangle \\ &= B \cdot \left( { x \wedge (x \cdot A) } \right),\end{aligned}$$ where we are able to use the very first step to write the scalar selection back in terms of dots and wedges.
The result of 4.154 probably requires similar logic.